Question

Electrolysis is carried out in three cells:

$\left(A\right)-1.0MCuS{O}_4$, Pt electrodes;
$\left(B\right)-1.0MCuS{O}_4$ , Copper electrodes;
$\left(C\right)-1.0MKCI$,Pt electrodes. If volume of

electrolytic solution is maintained constant in

each of the cells, which is correct set of pH

changes in (A),(B) and (C) cells respectively :

• A. increase in all the three
• B. decrease in all the three
• C. increase, constant, increase
• D. decrease, constant, increase

Answer 1

The correct option is D decrease, constant, increase

$\text{SRP of}C{u}^{2+}>\text{SRP of}{H}^{+}$
$\therefore$ In case 1)
copper gets reduced
and $O{H}^{-}$ gets oxidised
$\therefore \left[O{H}^{-}\right]$ decreases and $\left[H^{+}\right]$ increases,
$\therefore$ pH decreases $(pH=-log[H^{+}\left]\right)$

In case 2)
copper gets reduced and oxidised
$\therefore$ No change in conc of ions
Hence no change in pH

In case 3)
$\text{SRP of}{K}^{-1}<\text{SRP of}{H}^{+}$
$\therefore H^{+}$ gets reduced,
Hence $\left[H^{+}\right]$ decreases,
$\therefore pH=-log\left[H^{+}\right]$ increases